Real and complex behaviour of the base change function (was: The "cheta" function)

[jaydfox]

Can you make a picture for those that dont currently sit down with a computer algebra system computing exactly this?

A picture I can do.

Imho the \( \log^{[n-2]}(-1) \) converges to the upper primary fixed point of \( \exp \). So why should they come arbitrarily close to the real axis?

But Henryk, remember that, due to branching of logarithms, we must start at a point on the real line and then follow a path.

Well, if we start at \( \exp_{e}^{[n-2]}(x=0) \), you arrive at a rather large number. If you then try to create a simple path to \( \exp_{e}^{[n-2]}(x)=-1 \), and then perform n-2 logarithms, you will not get a simple path. It will loop wildly around the upper fixed point.

However, let's consider the point \( \exp_{e}^{[n-3]}(x)=0+\left(2\,\mathrm{floor}(\exp_{e}^{[n-3]}(0)/2\pi)+1\right)\pi i \).

This point is roughly the same distance from the origin as the [n-3] exponentiation of 0, but it has real part 0, and imaginary part equal to (2k+1)*pi, integer k. Thus, on the next exponentiation, it's -1.

Note that we can use a rather large quarter circle (or approximately so) to connect this point to the real line.

If we take the logarithm of this arc, in the primary branch, we get a line segment from the real line to a complex value with roughly the same real part, but imaginary part equal to pi/2.

Note that this line segment now gives us a very simple image as we iteratively perform logarithms, such that it ends up very, very close to 0, with a very small imaginary part (the higher the value of n, the closer to the real line it gets).

Now this approach only got us the "closest" singularity to whatever point on the real line we started at (0 in my example). Instead of a quarter circle, we can do a 3/4 circle, 5/4 circle, etc., and we can pick different k values in the (2k+1)*pi*i formula, to find arbitrarily many singularities, and depending on the exact path being used, we can find these singularities in arbitrarily many different branches.

I will draw up some pictures do demonstrate.